Midterm 3: Version E Answer Key

  1. \dfrac{\cancel{12}1\cancel{m^3}}{\cancel{5}1\cancel{n^2}}\cdot \dfrac{\cancel{15}\cancel{3}1\cancel{n^3}}{\cancel{36}\cancel{3}1\cancel{m^6}}\cdot \dfrac{\cancel{8}4m^{\cancel{4}}}{\cancel{6}3n^{\cancel{2}}}\Rightarrow \dfrac{4m}{3n}
  2. \dfrac{\cancel{x}1\cancel{(x+2)}}{\cancel{(x+2)}\cancel{(x+7)}}\cdot \dfrac{\cancel{2}1\cancel{(x+7)}}{\cancel{2}1x^{\cancel{3}2}}=\dfrac{1}{x^2}
  3. \phantom{1}
    \left(\dfrac{x-3}{7}-\dfrac{x-15}{28}=\dfrac{3}{4}\right)(28) \\
    \begin{array}{rrrrrrrrl} 4(x&-&3)&-&(x&-&15)&=&3(7) \\ 4x&-&12&-&x&+&15&=&21 \\ &+&12&&&-&15&&-15+12 \\ \midrule &&&&&&3x&=&18 \\ &&&&&&x&=&6 \end{array}
  4. \begin{array}{l} \dfrac{\left(\dfrac{x^2}{y^2}-36\right)y^3}{\left(\dfrac{x+6y}{y^3}\right)y^3}\Rightarrow \dfrac{x^2y-36y^3}{x+6y}\Rightarrow \dfrac{y(x^2-36y^2)}{x+6y}\Rightarrow \dfrac{y(x-6y)\cancel{(x+6y)}}{\cancel{(x+6y)}} \\ \\ \Rightarrow y(x-6y) \end{array}
  5. \begin{array}{l} \\ \\ \\ \\ \\ \\ \sqrt{x^6\cdot x\cdot y^4\cdot y}+2xy\sqrt{36\cdot x\cdot y^4\cdot y}-\sqrt{x\cdot y^2\cdot y} \\ \\ x^3y^2\sqrt{xy}+2xy\cdot 6y^2\sqrt{xy}-y\sqrt{xy} \\ \\ x^3y^2\sqrt{xy}+12xy^3\sqrt{xy}-y\sqrt{xy} \\ \\ \sqrt{xy}(x^3y^2+12xy^3-y) \end{array}
  6. \dfrac{\sqrt{7}}{3-\sqrt{7}}\cdot \dfrac{3+\sqrt{7}}{3+\sqrt{7}}\Rightarrow \dfrac{3\sqrt{7}+7}{9-7}\Rightarrow \dfrac{3\sqrt{7}+7}{2}
  7. \left(\dfrac{\cancel{x^0}1y^4}{z^{-12}}\right)^{\frac{1}{4}}\Rightarrow \dfrac{y^{4\cdot \frac{1}{4}}}{z^{-12\cdot \frac{1}{4}}}\Rightarrow \dfrac{y^1}{z^{-3}}\Rightarrow yz^3
  8. \phantom{1}
    (\sqrt{4x-5})^2=(\sqrt{2x+3})^2 \\
    \begin{array}{rrrrrrrr} &4x&-&5&=&2x&+&3 \\ -&2x&+&5&&-2x&+&5 \\ \midrule &&&2x&=&8&& \\ &&&x&=&4&& \end{array}
  9. \phantom{1}
    1. \left(\dfrac{x^2}{3}=27\right)(3)\Rightarrow x^2=81 \Rightarrow x=\pm 9
    2. \begin{array}{rrl} \\ \\ \\ 27x^2+3x&=&0 \\ 3x(9x+1)&=&0 \\ x&=&0, -\dfrac{1}{9} \end{array}
  10. \phantom{1}
    1. \begin{array}{rrl} \\ (x-12)(x+1)&=&0 \\ x&=&12, -1 \end{array}
    2. \begin{array}{rrl} \\ \\ x^2+13x+12&=&0 \\ (x+12)(x+1)&=&0 \\ x&=&-1, -12 \end{array}
  11. \phantom{1}
    \left(\dfrac{2}{x}=\dfrac{2x}{3x+8}\right)(x)(3x+8) \\
    \begin{array}{ll} \begin{array}{rrl} 2(3x+8)&=&2x^2 \\ 6x+16&=&2x^2 \\ 0&=&2x^2-6x-16 \\ 0&=&2(x^2-3x-8) \end{array} & \hspace{0.25in} \begin{array}{l} \\ \\ \\ \dfrac{-(-3)\pm \sqrt{(-3)^2-4(1)(-8)}}{2(1)} \\ \\ \dfrac{3\pm \sqrt{9+32}}{2} \\ \\ \dfrac{3\pm \sqrt{41}}{2} \end{array} \end{array}
  12. \begin{array}{rrl} \\ \\ (x^2-64)(x^2+1)&=&0 \\ (x-8)(x+8)(x^2+1)&=&0 \\ x&=&\pm 8 \end{array}
  13. \begin{array}{rrcrrrrrrrrrr} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ &&&&A&=&20&+&P&&&& \\ \\ &&L(L&-&5)&=&20&+&2(L&-&5)&+&2L \\ L^2&-&5L&&&=&20&+&2L&-&10&+&2L \\ &-&4L&-&10&&-10&-&4L&&&& \\ \midrule L^2&-&9L&-&10&=&0&&&&&& \\ (L&-&10)(L&+&1)&=&0&&&&&& \\ &&&&L&=&10,&\cancel{-1}&&&&& \\ \\ &&&&W&=&L&-&5&&&& \\ &&&&W&=&10&-&5&&&& \\ &&&&W&=&5&&&&&& \end{array}
  14. \phantom{1}
    x, x+2, x+4 \\
    \begin{array}{rrcrrrrrcrr} &&x(x&+&4)&=&35&+&10(x&+&2) \\ x^2&+&4x&&&=&35&+&10x&+&20 \\ &-&10x&-&55&&-55&-&10x&& \\ \midrule x^2&-&6x&-&55&=&0&&&& \\ (x&-&11)(x&+&5)&=&0&&&& \\ &&&&x&=&11,&-5&&& \end{array}
    \phantom{1}
    \text{numbers are }11,13,15\text{ or }-5,-3,-1
  15. \begin{array}{rrrrrrcrrr} \\ \\ \\ \\ \\ \\ &&&d_{\text{d}}&=&d_{\text{u}}&+&9&& \\ \\ &3(5&+&r)&=&4(5&-&r)&+&9 \\ &15&+&3r&=&20&-&4r&+&9 \\ -&15&+&4r&&-15&+&4r&& \\ \midrule &&&7r&=&14&&&& \\ &&&r&=&2&\text{km/h}&&& \end{array}

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Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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